How to compute asymptotic behavior of an inverse function? Let $f(x) := \frac{x}{1+\log_2(x)}$. I would like to compute an asymptotic expression for its inverse function - $f^{-1}(y)$ - when $y$ is large. For example: is $f^{-1}(y)$ is in $O(y \log y)$? Or in $O(y \log^k y)$ for some integer $k$?
I know how to compute asymptotics for functions, but how can I do the computation for inverse functions?
 A: This function is strictly increasing for $x \ge 2$ and things are relatively easy in this case, you can just plug in a guess and see how accurate it is to get bounds. More precisely:

Lemma: Let $f$ be a strictly increasing function. If $f(x) < y$ then $x < f^{-1}(y)$, and if $f(x) > y$ then $x > f^{-1}(y)$.

So for example we compute that
$$f(y \log_2 y) = \frac{y \log_2 y}{1 + \log_2 (y \log_2 y)} < \frac{y \log_2 y}{\log_2 y} = y$$
from which it follows that $f^{-1}(y) > y \log_2 y$. On the other hand
$$f(C y \log_2 y) = \frac{Cy \log_2 y}{1 + \log_2 (Cy \log_2 y)}$$
which we can see grows like $Cy$ and hence, for fixed $C > 1$, will be greater than $y$ for sufficiently large $y$. So for any $\varepsilon > 0$ we get that $f^{-1}(y) \le (1 + \varepsilon) y \log_2 y$ for sufficiently large $y$. This gives $f^{-1}(y) = O(y \log y)$ and in fact we get $f^{-1}(y) \sim y \log_2 y$. More precise asymptotics can be given in terms of the Lambert W function.
This particular example of asymptotic inversion occurs naturally in number theory, where it is used to go from the prime number theorem $\pi(n) \sim \frac{n}{\log n}$ to the asymptotic $p_n \sim n \log n$ for the $n^{th}$ prime number.
A: If you want the inverse of $$y= \frac{x}{1+\log_2(x)}$$ it is easier to rewrite it as
$$2y=\frac {2x}{\log_2(2x)}$$ or using natural logarithms
$$\frac 2 {\log(2)}y=\frac {2x}{\log(2x)}\implies x=-\frac{y}{\log (2)} W_{-1}\left(-\frac{\log (2)}{2 y}\right)$$ where $W_{-1}(.)$ is the second branch of Lambert function as @Qiaochu Yuan already answered.
In the real domain, this implies $y \geq \frac{e\log (2)}{2}$.
For large values of $y$, have a look at the expansion in the linked page
$$W_{-1}(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$  where $L_1=\log(-t)$ and $L_2=\log(-L_1)$.
Using only $L_1$, you then have $x \sim y \log_2(y)$ as @Qiaochu Yuan already answered.
A: Informally: from $$y=f(x)= \frac{x}{1+\log_2(x)} \tag 1$$
you first see that $f(x)$ is increasing for $x>2$, hence you are interested in both $x,y\to +\infty$  (actually, there is also another branch, where $y \to \infty$, for $x \to 1/2$, but I assume you are not interested in that one).
Then, for large $x$ the one in the denominator becomes negligible and
$$ x \approx \tag 2 y \,\log_2(x)$$
Now, you can expect that $(2)$ can be applied recursively, so
$$ x \approx y \log_2( y \log_2(x)) \approx y \log_2( y \log_2(y \log_2(\cdots))) \tag 3$$
or
$$ x = y (\log_2 y +O(\log_2 \log_2 y)) \tag 4$$
This is a mere guess, but you can make it rigorous.
For example, it's easy to show, plugging from $(1)$, that
$$ \frac{x}{y \log_2 y} \to 1$$
and
$$ \frac{x}{y} - \log_2 y \approx 1 + \log_2(1+\log_2(x)) \to +\infty$$
